A subset $\mathcal{C}\subseteq\{0,1,2\}^n$ is said to be a $\textit{trifferent}$ code (of block length $n$) if for every three distinct codewords $x,y, z \in \mathcal{C}$, there is a coordinate $i\in \{1,2,\ldots,n\}$ where they all differ, that is, $\{x(i),y(i),z(i)\}$ is same as $\{0,1,2\}$. Let $T(n)$ denote the size of the largest trifferent code of block length $n$. Understanding the asymptotic behavior of $T(n)$ is closely related to determining the zero-error capacity of the $(3/2)$-channel defined by Elias'88, and is a long-standing open problem in the area. Elias had shown that $T(n)\leq 2\times (3/2)^n$ and prior to our work the best upper bound was $T(n)\leq 0.6937 \times (3/2)^n$ due to Kurz'23. We improve this bound to $T(n)\leq c \times n^{-2/5}\times (3/2)^n$ where $c$ is an absolute constant.

翻译：子集 $\mathcal{C}\subseteq\{0,1,2\}^n$ 被称为（块长度为 $n$ 的）**三进制不同码**，若对任意三个不同码字 $x, y, z \in \mathcal{C}$，存在某个坐标 $i\in \{1,2,\ldots,n\}$ 使得三者在该坐标上全部不同，即 $\{x(i),y(i),z(i)\}$ 等于 $\{0,1,2\}$。设 $T(n)$ 表示块长度为 $n$ 的最大三进制不同码的规模。理解 $T(n)$ 的渐近行为与确定 Elias'88 定义的 $(3/2)$-信道的零错误容量密切相关，且是该领域长期悬而未决的公开问题。Elias 曾证明 $T(n)\leq 2\times (3/2)^n$，而在此工作之前的最佳上界是 Kurz'23 给出的 $T(n)\leq 0.6937 \times (3/2)^n$。我们将该界改进为 $T(n)\leq c \times n^{-2/5}\times (3/2)^n$，其中 $c$ 为绝对常数。